CQU ENTA11005 Assignment 1 Solutions

# CQU ENTA11005 Assignment 1 Solutions - ENTA 11005...

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ENTA 11005 – Engineering Science 1 of 7 Assignment 1 Solutions Question 1 (3 Marks) (a) We need to calculate the volume conversion from pints/pots to m 3 L mL cm L m cm m m 1 1000 1000 1 1 100 1 1 3 3 3 3 × × × = mL mL mL m 6 3 3 10 1 100000 100 1 1 × = = × = Knowing that 1m 3 is equivalent to 1×10 6 mL, we can determine how many pints/pots are in 1m 3 ; For Paddy ; No. of pints in 1m 3 4 . 1754 570 10 1 570 1 6 3 = × = = mL mL mL m For Steve ; No. of pots in 1m 3 8 . 3508 285 10 1 285 1 6 3 = × = = mL mL mL m So Paddy will be drinking 1754 pints, and Steve will have 3509 pots throughout the year. (b) To find the mass ; V m = ρ , or V m × = where ρ is density, m is mass and V is volume. ρ = 1.35 kg / m 3 3 4 6 3 10 7 . 5 10 1 1 570 m mL m mL V × = × × = ( ) ( ) 6555 . 0 10 7 . 5 10 15 . 1 4 3 = × × × = × = V m The mass of one pint is 0.656 kg , or 656 g . To find the weight ; mg W = , where m is mass and g is acceleration due to gravity.

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ENTA 11005 – Engineering Science 2 of 7 m =0.6555kg g = 9.8 ms -2 4239 . 6 8 . 9 6555 . 0 = × = W The weight of one pint is 6.4 N . (c) The mass will be exactly the same on the moon as it is on earth (or anywhere in the universe!) So mass is 656g. Weight on the other hand is the force with which an object experiences due to gravity. Because the acceleration due to gravity is much smaller on the moon than it is on earth, the weight of the pint will be considerably less. 09385 . 1 67 . 1 6555 . 0 = × = = moon moon mg W So the weight of the pint on the moon is 1.09 N .
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## This note was uploaded on 03/25/2008 for the course ENTA 11005 taught by Professor Colingreensill during the Spring '08 term at A.T. Still University.

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CQU ENTA11005 Assignment 1 Solutions - ENTA 11005...

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